# How do you find the vertex and intercepts for  f(x) = -3x^2 + 5x + 5?

Vertex is at $\left(\frac{5}{6} , \frac{85}{12}\right)$ Y intercept is at (0,5) and X intercepts are at (-0.703,0) and (2.37,0).
Here a=-3 ; b=5 ; c=5 we know the co-ordinate ${x}_{1}$ of the vertex is equal to (-b/2a). $\therefore$ ${x}_{1} = - \frac{5}{-} 6 = \frac{5}{6}$. Now Putting the value of x -cordinate in the equation we get $y = - 3 \cdot {\left(\frac{5}{6}\right)}^{2} + 5 \cdot \frac{5}{6} + 5$ or $y = \frac{85}{12}$ $\therefore {y}_{1} = \frac{85}{12}$ $\therefore$ Vertex is at $\left(\frac{5}{6} , \frac{85}{12}\right)$
Now to find Y-intercept putting x=0 we get y = 5 i.e The parabola cuts the Y axis at 5. To get X intercepts putting y=0 ; we get the equation as $- 3 {x}^{2} + 5 x + 5 = 0$ Solving for x in the above quardratic equation we get two roots of x as $- \frac{5}{2 \cdot \left(- 3\right)} + \left(\frac{\sqrt{{5}^{2} - 4 \cdot \left(- 3\right) .5}}{2 \cdot \left(- 3\right)}\right)$ and$- \frac{5}{2 \cdot \left(- 3\right)} - \left(\frac{\sqrt{{5}^{2} - 4 \cdot \left(- 3\right) .5}}{2 \cdot \left(- 3\right)}\right)$ Which gives the two roots as $- 0.703 \mathmr{and} 2.37$ So The parabola cuts X-axis at $- 0.703 \mathmr{and} 2.37$ points. [Answer]